What are the solutions to the equation (x-5)(x+2)=18 ?

The solutions to the equation (x-5)(x+2)=18 are x=7,x=-4

Find the zeros of (x-5)(x+2)=18

The zeros of (x-5)(x+2)=18 are x=7,x=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x - 5) (x + 2) = 18

x = 7, x = - 4

Solution steps

(x - 5) (x + 2) = 18

Expand

(x - 5) (x + 2) : x^{2} - 3 x - 10

x^{2} - 3 x - 10 = 18

Move

18

to the left side

x^{2} - 3 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 28) = 11

x_{1, 2} = \frac{- ( - 3 ) \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 11}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 11}{2 \cdot 1}

x = \frac{- ( - 3 ) + 11}{2 \cdot 1} : 7

x = \frac{- ( - 3 ) - 11}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 7, x = - 4

(x - 13)^{2} = (x - 12)^{2} + (x - 5)^{2}

(x) = x^{2} - 160 x + 10000

x, z = 9 x^{2} + 18 x y + 6 y^{2}

4.9 x^{2} - 100 x - 9 = 0

x, f (x + h) = 3 (x + h)^{2} + 2 (x + h) - 4

What are the solutions to the equation (x-5)(x+2)=18 ?
The solutions to the equation (x-5)(x+2)=18 are x=7,x=-4
Find the zeros of (x-5)(x+2)=18
The zeros of (x-5)(x+2)=18 are x=7,x=-4